# Simple computational methods for the analysis of (s, s) type inventory models

## Full text

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### Department of Statistics, Annamalai University, Annamalainagar, Tamilnadu, India

ARTICLE INFO ABSTRACT

A simple computational method (SCM) to analyze a class of (s, S) type inventory problem is developed. Under this (s, S) policy, (i) the number of units demanded where d = 1, 2… a ( s) at successive demand epochs form a Markov chain (MC) with one step transition probability matrix (TPM) P and (ii) the replenishments are instantaneous. This method gives the algorithm for computation of stationary probabilities of inventory process, joint probability function of number of transitions and quantities of replenishments per cycle, conditional and unconditional average costs. Illustrative example for a few special cases are provided, which strengthen the applicability of the SCM to practical problems.

### INTRODUCTION

Numerous methods have been suggested by various authors for formulating and finding optimal inventory policies which have more number of practical applications in real life situations. Such eventualities with modifications have been increasing in the inventory policies of the (s,S) types also. One more possible way out is discussed below. In this paper, a different method of approach for the analysis of (s, S) inventory models, wherein the number of units demanded at successive demand epochs are Markov dependent as introduced by Krishnamoothy and Lakshmi (1991), is discussed. For the detailed discussion

*Corresponding author:maalinga@gmail.com

about the systematic analysis of various types of (s, S) inventory system refer Srinivasan and Ravichandaran (1994), Hiller and Liberman, (1990) and Krishnamoorthy et al. (1995).

A Markov Dependent [(s, S), a, P] Inventory Model

For the present (s, S) inventory problem, it is assumed that the bulk quantity demanded, say ‘d’ at a demand epoch is a random variable with d = 1, 2, … , a ; a  s such that the sequence of units demanded in the successive demand epochs forms a finite state Mc on the state space {1, 2, …, a } with TPM ‘P’. To avoid the perpetual shortage, it is assumed that (S-s) > s. Further if the inventory level is s+1 or greater at a demand epoch, then no order is placed. On the other hand, if the inventory

### 0975-833X

International Journal of Current Research Vol. 33, Issue, 3, pp.060-067, March, 2011

INTERNATIONAL JOURNAL OF CURRENT RESEARCH

Article History:

Received in revised form 15th January, 2011

Accepted 7th February, 2011

Published online 13th March, 2011

Key words:

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level either falls to ‘s’ or dips below ‘s’ at an epoch, an order is placed for at least (S-s) units. The quantity ordered is subject to review at the epoch of replenishment so as to bring inventory to level S. Zero lead time is assumed. This kind of inventory model may be represented in symbols as [(s, S), a, P]. Further the inter occurrence times between successive demand epochs are assumed to be independent and identically distributed (i.i.d) random variables with distribution function G (.) and density function g (.). If the demand is for ‘i’ units at an epoch, it may be called as i-type demand for I = 1, 2… a.

Notations

{I (t)} : inventory level process. * : Convolution

E1 : {1, 2… a-1, a}

E 2 : {s+1, s+2… s-1, S}

E :

         

         

    

 

 

 

 

 

 

2 s a, , 2 S 1, a , , 2 s 3, , 2 s 2, , 2 s 1,

1 s a, , 1 S 1, a , , 1 s 3, , 1 s 2, , 1 s 1,

 

1,S-a

, 2,S-a2

, 3,S-a 1

, ,

a 1,S-a 1

### 

,

a -S a, , a -S 1, a , , a -S 3, , a -S 2, , a S 1,

  

 

 

2) S (2, 2) S (1,

3) S (2, 3) S (2, 3) S (1,

 

 

     

S) (a, ,

(3.S), S), (2, S), (1,

1) S (1,

E3 : {S-s, S-s+1 … S-s+ (a-1)}

E4 : {s-a+1, s-a+2… s-1, s}

E5(r) : {r, r+1… S-s}, where r is the smallest

integer

### 

a

s S 

N : {1, 2…}

Here, the set E is a proper subset of E1  E2 i.

e., E  E1  E2. It is due to the fact that the

continuous time, inventory process decreases from the levels S, S-1, S-2 … etc, with instantaneous reflecting barriers at each of its levels i  E4

depending upon the demands. Hence the inventory level never decreases to zero at any instant. Further E3 forms the set of all possible number of units of

replenishments. The set E5 (r) is the collection of

all possible number of transitions in a cycle of replenishments.

Markov renewal inventory process

Let 0 = T0 < T1 < T2 … be the successive demand

epochs and X0, X1 , X2, … be the number of units

demanded at these epochs respectively, so that the sequence X = { Xn } forms a MC on the states i,

j  E1 with initial distribution p [ X0 = i ] = Pi, and

transition probabilities

1

### P

n1 n ij (1)

Assume that the TPM P = (Pi j) is irreducible and

a periodic throughout the discussions to follow. Let (S =) Y0, Y1, Y2 … represent the inventory

levels just after meeting the demands at T0, T1, T2,

…, so that time intervals between successive occurrences of bulk demands induce the triple sequence process { ( X, Y ), T } = { ( Xn , Yn ), Tn,

n  N } as a Markov Renewal inventory process (MRIP) on the state space E with the Semi- Markov kernel Q (t) = [ {Q ( i, I ) , (j, J) ; t } ] where Q { (i. I), (j, J); t)}

=P

Xn1j,Yn1J

; Tn1Tnt

Xni,YnI

### 

P(i,I), (j.J) G(t) 2)

The TPM of the underlying (X, Y) double

sequence MC is given by Q () =

### Q

((i, I), (j, J)) (say).

Stationary Distribution of the (X, Y) MC

Let the stationary probability vector of the (X, Y) MC denoted by the row vector

 i,j

## 

for I = 1, 2 … a,

j = s+1, s+2… S-a, i = 1,2, …,n, J = S-n ; n = a-1, a-2, …, 1

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unique solution vector as the state space E is finite dimensional and P is irreducible. A Simple Computational Method (SCM) is capsule below as an algorithm.

STEP- I: Using the TPM P = (PiJ), I, J  E1 obtain

its stationary probability vector, say  = (1, 2 …

a) by solving p =, e = 1

STEP-II: Arrange the states as they are ordered in

the set E of section 1.1 and identify the TPM

### Q

of the (X, Y) MC in terms of PiJ’s and zeros.

STEP- III: Omit the last set ‘of ‘a’ equations corresponding to  (J,S), J = 1,2, … a , from

Q Π

 . Then select the remaining stationary equations together with

### 

2

j

i j) (i, v

π

E

so that the

resulting independent set of simultaneous equations is solvable.

STEP- IV: Make suitable changes such that one set of ‘a’ equations corresponding to

1,S 1

2,S 2

a,S a

 

### 

 appearing on the

left of the system

### Q

reduce to a system of ‘a’ equations of the form

     

## 

1,S1 2,S2 a,Sa

### 

Where A

is a full rank square matrix of order of ‘a’. Hence obtain values for

1,S1

2,S2

### π

a,Sa

by any one of the standard methods.

STEP-V: Using the values of

1,S 1

2,S 2

a,S a

 

### 

 and selecting

expressionsfor

1,S1,π2,S2, π2,Sa π2,Sa-1

π     and

and compute the next set of ‘a’ components in the

solution of

### Q

. Continue it, until values for all components

 i,j

### E

of the II vector

are obtained.

Mean Sojourn times

For the MPIR {(X, Y), T}, let the mean sojourn time at state

 i,j

### E

be  (i, J). Since the

inter occurrence times between transition epochs are assumed to be i.i.d random variables, with d.f G (.), it is seen that

 

### 

 

0 j

i, 1 G(t) dt μ

μ (4)

Inventory Level Process {I (t)}

Since I (t) gives the unhand stock level at time ‘t’, i.e.,

(t)

n

n

### T

n1 it is seen that {

I (t) ; t  0 } is a semi -Markov Process on the state space E2 with Kernel Q Let

## 

(t)

0

### 

Then by the application of key renewal theorem. It can be established that

t

### 

exists. Thus using the

results of (4), it can be shown that for i  E1

### 

 

 

a

1 j

n) (j, a

k) (1,

k) (1, k) (1, a

1 j

n) (j, n) (j,

π μ

π μ π P(n)

1 E

(5a)

where n = s+1, s+2,….,S-a and n = S

### 

 

 

n -S

1 j

n) (j, a

k) (1,

k) (1, k) (1, n -S

1 j

n) (j, n) (j,

π μ

π μ π P(n)

1 E

(5b)

where n = S-a+1, S -a+2,….,S-1

Expected Total Costs

In this type of [(s,S), a, P] inventory situations, demand cum replenishment epochs play a very crucial role in obtaining cost expressions, Hence to study the characteristics of the time durations of the successive replenishment epochs Wn and Wn+1,

n  N under steady state conditions. Assume that an i-type demand occurs at Wn while a J -type

demand takes place at Wn+1. Let is be called as

J-cycle. Further assume that ‘M’ denotes the quantity of replenishment at Wn+1, Z = (Wn+1- Wn) denotes

the random length of the cycle between the epochs Wn and Wn+1 and K denotes the random number of

demand epochs in it with the K-th one being the last demand epoch of the cycle i.e. at Wn+1.

Then materialization of the events (M = m), (K = k) and z  (z, z + dz) implies that m  E3, K  E 5

(r) and z  [0,). Hence for a given i, J  E1,

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(i) S1 = sum of the demands of the first

(k-1) demands in the J-cycle is strictly less than (S-s),

(ii) S2 = sum of the demands of all

k-demand epochs is equal to m and (iii) z  Z  z+ dz is to take the

following form

P ,P , ,P , ,g

###  

z dz

B 1 dz z k; m;

f *k

i i i i, i i, j

i,  1 2  k2 k

Where i, i1, i2, , ikE1, S1

Ss

### 

S2 = m, m  E3, K  E 5 (6)

and B is the normalizing constant and ik = j. From

(6) with little effort one could obtain the marginal distribution for M, K and Z. The joint probability function (Pf) of M and K is sufficient to derive the cost per transition and is given by

, P , , P , P B 1 k) (m, f

1 1 k 2 1 i,i i ,j

i i,

j 

### 

 

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i, i1, i2, …, ik E 1, s1 < S-s and S2 = m, m E 3

, K  E 5 since iK = J. For the constructional details

of

j

### .

refer appendices I (a), II (a), II (b), III (a)

for some specific cases

Marginal Distributions in J-cycle

Let the marginal probability functions of quantity of replenishment M, number of transitions K in a J-cycle be respectively designated by

j(1)

and

(1)

j

Thus

 

5

k

1 j

(1)

j (m) f(m,k); j

f

E

E

 

5

k

1 j

(2)

j (k) f(m,k); j

f

E

E

Further let

### f

j* = probability that a cycle is

terminated by a J-demand.

=

### 

5 k 5

j m

k) (m, f

E E

Hence, the following are the conditional expected values of the random variables (m/K), K and M in the J-cycle.

f(m,k)

k m K

M E

3 5

E m kE

j

 

     

    

(8a)

### 

5

k

(2) j j(K) k f (k) E

E

(8b)

### 

3

m

(1) j j(K) m f (m) E

E

(8c)

Average Inventory per Transition

In a J-cycle, choose a demand epoch Tn at random,

Let (j, i)  E be the state of the (Xn, Yn) process at

Tn. The inventory level during the transition

interval (Tn -Tn-1) is (j + i), if I = S and it is equal

to [S - M + j ] if i = S. Further let I+= I (

### T

n) be the inventory level just after the demand epoch Tn.

Hence the conditional expected inventory level during the interval (Tn-Tn-1), given that (Xn,Yn) =

(j,1) is given by

j

### 

(j,S)

i) , j S) (i

j(I ) (j i)π S E(M) jπ

E

2 i

   

### 

 

E

( (9)

Cost Aspects

Consider the cost considerations for a j-cycle as Hj

the unit storage cost per transition Lj fixed ordering

cost Cj unit variable cost. Then the conditional

expected total cost per transition (TCT) in a j - cycle is given by

say , T ) (I E H K M E C (k) E

L (TCT)

E j j j j j

j j

j  

     

  (10a)

Thus the expected total cost per transition in any cycle is given by

### 

1

j * j jf

T E(TCT)

E

(5)

Expected Long Run cost Per Unit time

For a given J  E1, the expected long run total cost

(TC) per unit time in a j-cycle is given by

MZ

h E(I) E

C (z) E

L (TC)

E j j j

j j

j   

(11a)

Thus

a 1 j

j j(TVC)f

E

E(TC) (11b)

where

2 n

p(n) n E(TC)

E

###   

        

0

3 5

m k

j

j (m / z) f (m,k;z) dz

Z M E

E E

and

###   

 

0

3 5

m k j

j Z z f (m,k;z) dz

E

E E

OPTIMIZATION PROBLEMS

The best policies under two different contexts are worked out assuming specific values for the parameters involved. These are supported by numerical illustrations.

Example 1

It is assumed that a retailer keeps stock pile according to the [(S, s), a, P] system. The best policy given the values s = 2, a = 2, P11 = .3, P12 =

.4, L1= 150 L2= 200 C1 = C2 = 50, H1 = H2 = 10

when S = 5, 6 and 7 is determined by using the calculations of tables 1(a), 1(b) and 1(c) given below which are outlined in expressions (1) - (10) and appendices (I) through (III). Let P0 =

        

.6 .4

.7 .3 P0

The resulting calculation of E (TCT) for three values 5,6,7 of S is given in Table 1(c) which indicates that the over all minimum E (TCT) corresponds to S = 7. Hence [(2, 7), 2, P0] is the

most desirable policy, among the above three [(2, S), 2, P0]; S = 5, 6, 7 policies.

Example 2

It is assumed that the demand for the product is seasonal according to three different TPM’s obtained on the basis of the historical data. Now the problem is to identify the season which has the minimum E (TCT). For demonstration, [(2, 7), 2, P] policy is again investigated at three different seasons I, II and III respectively with P = P1

=

, P = P2 =

, P = P3

=

### .9

. Further, placing s =2, a = 2, L1 =

100, L2 = 150, C1 = C2 = 35, H1 = H2 = 8 the

corresponding results are computer, as reported in From this table, It may be observed that the season (*) corresponds to the P matrix which yields the highest v1 value. It may be due to the fact that the

number of transition in a cycle increases with increasing v1 values and this may minimize the

### 

*

j j j j

j j

j E(K) C E(M / K) H E(I ) f

L

E(TCT)

### 

because number of transitions ‘K’ in a ‘j’ cycle appears in the denominator of the first two terms in the above E (TCT) expression.

APPENDIX I (a)

Joint PF of M and K for [(s, S), a, P] = [(2, 7), 2, P] model

  

   

 

 

 

  

 

5 k )P

P (P

4 k P P ) P (P

) P P P P (P ) P (P

3 k P

P )

P (P

B 1 k) (5, f

4 11 21 11

2 11 21 21 11

21 12 11 21 12 21 11

21 22 22

12

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Table 1(a). ij values for [ (s, S), a, P] = [ (2,7), 2, P0] model

j 3 4 5 6 7

i .3 .4 .5 .6 .7 Ei (I+)

1 .0576 .0858 .0342 .1139 .0722 2.1525 2. .1372 .0797 .1889 0 .2306 3.5927 p(n) .1948 .1655 .2231 .1139 .3028 5.7452

jP(n) .5844 .6620 1.1155 .6834 2.1196 5.1649

Table 1(b).

j

### k)

Values for [(s, S), a, P] = [(2, 7), 2, P0] model

3 4 5 fj(m) Ej(M) Ej(M/K) *

j

### f

j=1 .1560 .0822 .0029 .2411 1.2055 .3657

### f

(2)

1

.1560 .0822 .0029 .2411

### k.f

(2)

1

.468 .3288 .0145 .8113 = 1E (k)

j=2 .3290 .0221 0 .3511 4.2029 1.3028

.2340 .1672 .0066 .4078

### f

2(2) .5630 .1893 .0066 .7589

### k.f

2(2) 1.689 .7572 .0330 2.4792 = E2 (k)

Table 1(c). E (TCT) values for [(s, S), a, P] = [(2, S), 2, P0] models when S = 5, 6, 7

S J vj Lj/Ej(k) CjEj(M/k) HjEj(I) Tj * j

j

### f

j* E(TCT)

5 1 .36 244.10 21.08 17.83 283.01 .29 82.07 242.72 2 .64 134,18 62.26 29.83 226.27 .71 1260.6

6 1 .36 265.02 12,21 19.45 296.68 .19 56.37 219.46 2 .64 96.36 72.69 32.29 201.34 .81 163.09

7* 1 .36 184.89 9.15 21.53 215.57 .24 51.74 165.09 2 .64 80.65 32.57 35.93 149.15 .76 113.35

Table 2. E (TCT) values for [(2, 7), a, P] = [(2, 7), 2, P] model when P varies

S J vj Lj/Ej(k) CjEj(M/k) HjEj(I) Tj *

j

j

### f

j* E(TCT) I 1 .36 123.26 12.80 17.22 153.28 .24 36.79 139.2

2 .64 60.49 45.60 28.74 134.83 .76 102.4

II 1 .50 78.03 16.59 22.99 117.69 .34 39.99 123.4 2 .50 65.81 37.24 23.47 126.52 .66 83.50

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               5 k 0 4 k P P P ) P (P 3 k P P ) P (P P P ) P (P B 1 k) (5,

f 11 21 11 11 12

22 12 21 11 12 21 22 12 2                   4 12 11 21 22 12 22 12 11 12 21 12 21 11 22 22 12 2 P)P (P 4 k P P )P P (P ) P P P P P (P ) P (P 3 k P ) P (P B 1 k) (6, f where       

### 

   5 3 k j 6 5 m 2 1 j k) (m, f 1 B

APPENDIX II (a)

Stationary Distribution of [(s, S), a P] = [(2, 6), 2, P] model STEP-I: ) P (P P ν and ) P (P P ν to leads 1 νe ν,

νP  1 21 12 21 2 12 12 21

STEP-: TPM of

### Q

is found as to

From (1, 3) (2, 3) (1, 4) (2, 4) (1, 5) (1, 6) (2, 6)

                            0 0 P P 0 0 0 0 0 P P 0 0 0 0 0 0 0 P P 0 P 0 0 0 0 0 P P 0 0 0 0 0 P P P 0 0 0 0 0 P P 0 0 0 0 0 (2,6) (1,6) (1,5) (2,4) (1,4) (2,3) (1,3) 21 22 11 12 11 12 22 21 12 11 22 21 12 11

STEP – III

P11 (1.4) + P21 (2.4) = (1.3)

(1)

P12 (1.5) = (2,.3)

(2)

P11 (1.5) = (1.4)

(3)

(1.3) + (1.4) + (1.5) + (1.6) = v1 (4)

(2.3) + (2.4) + (2.6) = v2 (5)

((1,6), (2,6)) P = ((1,5), (2,4)) (6)

STEP – IV

From (2) and (3), it is formed that

### 

         0 0 P P π , π π ,

π(1,4) (2,3) (1,5) (2,4) 11 12

(7)

(1,5)

(2,4)

## 

P on both sides of (6),

(1,5)

(1,6)

(2,4)

(2,6)

(1,5)

(2,4)

### P)

(8) Adding (8) with (7) P, and using the facts (4) and (5), it is obtained that

                              

 P

0 P 0 P P 0 0 P P P I π , π v 21 11 12 11 (2,4) (1,5)

APPENDIX II (b)

Joint PF of M and K for [(s, S), a, P] = [(2, 6), 2, P) model             P P P ) P (P P P ) P (P P P ) P (P 0 B 1 k) (4, f 11 11 11 21 11 11 21 21 11 21 12 21 11 1              4 k 0 3 k P P ) P (P 2 k P )P P (P B 1 k) (5,

f 11 21 11 12

11 21 21 11 5             P P P P) (P P P ) P (P P P ) P (P 0 B 1 k) (5,

f2 12 22 21 12 11 21 12 22

where

     

### 

   4 2 k j 5 4 m 2 1 j k) (m, f 1 B

APPENDIX III (a)

Stationary Distribution of [(s, S), a P] = [(2, 5), 2, P] model STEP-I: ) P (P P ν and ) P (P P ν to leads 1 νe ν,

νP  1 21 12 21 2 12 12 21

STEP- : TPM of

### Q

is found as To

(8)

     

 

     

 

0 0 0

0 0 0

0 0 0

0 0 0

21 22

11 12

12 11

22 21

12 11

P P

P P

P P

P P

P P 0 0 0

(2,5) (1,5) (1,4) (2,3) (1,3)

STEP - III: Selection of stationary equations

from

## 

### Π

, e = 1 excepting the last set of

((1.5) , (2.5)) P = ((1.4) , (2.3))

(1)

P11(1.4) = (1,.3)

(2)

(1.3) + (1.4) + (1.5) = v1

(3)

((2.3) + (2.5)) = v2 (4)

(1,4) (2.3) + (2.5)

P = ((1,4), (2,3)) (I + P) (5)

Rearrangement of (2) gives

### 

   

    

0 0

0 P π , π ,0

π(1,3) (1,4) (2,3) 11 (6)

Addition (6) with (5) and using (3) and (4), it is simplified as

### 

    

  

         

 P

0 0

0 P P I π , π

vP (1,4) (2,3) 11

(7)

(1,4)

(2,3)

## 

### that

δ δ π and δ δ

π(1,4)1 (2,3)2

where A

and , A A

A A

22 21

12 11

 1, 2,  are the

determinant values with If

A A

A A δ , v v

A A δ , A A

v v δ

22 21

12 11

2 1

12 11 2 22

21 2 1

1 

### REFERENCES

Hiller, F.S and Liberman, G.J. 1990. Introduction to Operation Research, Fifth-edition, McCraw Hill.

Krishnamoorthy, A and Varghese, T.V. 1995. On a Two-commodity Inventory Problem with a Markov-Shift in demand, OPSEARCH, Vol.32, No.1, p44-56.

Krishnamoorthy, A and Lakshmi. B. 1991. An inventory model Markov dependent demand quantities, Cahiers,du, CERO Vol.33, No.1-2, 91-101.

Sivasamy. R and Senthamaraikannan. K 1994. Embedded processes of a Markov Renewal Bulk Service Queue, Asia Pacific Journal of Operation Research, 11-51.

Srinivasan. S.K and Ravichandran. N. 1994. Multi-iterm (s, S) inventory model with Poisson demand, General lead time and adjustable Recorder time. Proceedings of the international Conference on Stochastic Models, Optimization techniques and computer (p.226-237) applications, Willey Eastern Ltd.,,

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